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2 years ago

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A wave travels through a medium at 251 m/s and has a wavelength of 5.10 cm. What is its frequency? What is its angular frequency

? What is its period of oscillation?

1 answer:

allochka39001 [22]2 years ago

5 0

Explanation:

It is given that,

Speed of a wave, v = 251 m/s

Wavelength of the wave, λ = 5.1 cm = 0.051 m

(1) The frequency of the wave is given by :

$\nu=\dfrac{v}{\lambda}$

$\nu=\dfrac{251\ m/s}{0.051\ m}$

$\nu=4921.56\ Hz$

(2) Angular frequency of the wave is given by :

$\omega=2\pi\nu$

$\omega=2\pi\times 4921.56\ Hz$

$\omega=30923.07\ rad/s$

(3) The period of oscillation is given by T as :

$T=\dfrac{1}{\nu}$

$T=\dfrac{1}{4921.56}$

T = 0.000203 seconds

or

T = 0.203 milliseconds

Hence, this is the required solution.

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A 0.300kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.800m/s when it makes a head-o

e-lub [12.9K]

Answer:

The final velocity of the first glider is 0.27 m/s in the same direction as the first glider

The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

0.010935 J

0.0858675 J

Explanation:

$m_1$ = Mass of first glider = 0.3 kg

$m_2$ = Mass of second glider = 0.15 kg

$u_1$ = Initial Velocity of first glider = 0.8 m/s

$u_2$ = Initial Velocity of second glider = 0 m/s

$v_1$ = Final Velocity of first glider

$v_2$ = Final Velocity of second glider

As momentum and Energy is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

${\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}$

From the two equations we get

$v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.3-0.15}{0.3+0.15}\times 0.8+\frac{2\times 0.15}{0.3+0.15}\times 0\\\Rightarrow v_1=0.27\ m/s$

The final velocity of the first glider is 0.27 m/s in the same direction as the first glider

$v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.3}{0.3+0.15}\times 0.8+\frac{0.3-0.15}{0.3+0.15}\times 0\\\Rightarrow v_2=1.067\ m/s$

The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

Kinetic energy is given by

$K=\frac{1}{2}m_1v_1^2\\\Rightarrow K=\frac{1}{2}0.3\times 0.27^2\\\Rightarrow K=0.010935\ J$

Final kinetic energy of first glider is 0.010935 J

$K=\frac{1}{2}m_2v_2^2\\\Rightarrow K=\frac{1}{2}0.15\times 1.07^2\\\Rightarrow K=0.0858675\ J$

Final kinetic energy of second glider is 0.0858675 J

6 0

2 years ago

If a young protostar with a disk is rotating and shrinking. how much faster is it rotating after its size has decreased by a fac

maks197457 [2]

In this system we have the conservation of angular momentum: L₁ = L₂
We can write L = m·r²·ω

Therefore, we will have:
m₁ · r₁² · ω₁ = m₂ · r₂² · ω₂

The mass stays constant, therefore it cancels out, and we can solve for ω<span>₂:
</span>ω₂ = (r₁/ r₂)² · ω<span>₁

Since we know that r</span>₁ = 4r<span>₂, we get:
</span>ω₂ = (4)² · ω<span>₁
= 16 </span>· ω<span>₁

Hence, the protostar will be rotating 16 </span><span>times faster.</span>

5 0

2 years ago

An object has a position given by r = [2.0 m + (2.00 m/s)t] i + [3.0 m − (1.00 m/s^2)t^2] j, where quantities are in SI units. W

lidiya [134]

Answer: 1 m/s

Explanation:

We have an object whose position $r$ is given by a vector, where the components X and Y are identified by the unit vectors $i$ and $j$ (where each unit vector is defined to have a magnitude of exactly one):

$r=[2 m + (2 m/s) t] i + [3 m - (1 m/s^{2})t^{2}] j$

On the other hand, velocity is defined as the variation of the position in time:

$V=\frac{dr}{dt}$

This means we have to derive $r$ :

$\frac{dr}{dt}=\frac{d}{dt}[2 m + (2 m/s) t] i + \frac{d}{dt}[3 m - (1 m/s^{2})t^{2}] j$

$\frac{dr}{dt}=(2 m/s) i - (\frac{1}{2} m/s^{2} t) j$ This is the velocity vector

And when $t=2s$ the velocity vector is:

$\frac{dr}{dt}=(2 m/s) i - (\frac{1}{2} m/s^{2} (2 s)) j$

$\frac{dr}{dt}=2 m/s i - 1m/s j$ This is the velocity vector at 2 seconds

However, the solution is not complete yet, we have to find the module of this velocity vector, which is the speed $S$ :

$S=\sqrt {-1 m/s j + 2 m/s i}$

$S=\sqrt {1 m/s}$

Finally:

$S=1 m/s$ This is the speed of the object at 2 seconds

6 0

2 years ago

During the class prize-giving ceremony, Anand clapped his hands hard while Kumar clapped his hands softly. Everybody could hear

algol13

Answer:

C - higher volume

Explanation:

The pitch or frequency of sound that an object can produce depends upon its size and configuration . The shape of hand of all are same so the frequency of sound produced by hands of all will be almost same . Hence frequency of sound produced by the hands of Anand and Kumar would have been almost the same .

But the intensity of sound produced by them would have been different . Intensity represents energy a sound carries . Hard hitting clap will produce sound of higher intensity . Intensity of sound is also called high volume sound . So Kumar's clap will carry greater energy and hence greater volume of sound .

3 0

2 years ago

the water behind hoover dam in nevada is 206 m higher than the colorado river below it. at what rate must water pass through the

Nitella [24]

Answer:

the required mass flow rate is 49484.37 kg/s

Explanation:

Given the data in the question;

we first determine the relation for mass flow rate of water that passes through the turbine;

so the relation for net work on the turbine due to the change in potential energy considering 100% efficiency is;

$W_{net}$ = m ( Δ P.E )

so we substitute (gh) for ( Δ P.E );

$W_{net}$ = m (gh)

m = $W_{net}$ / gh

so we substitute our given values into the equation

m = 100 MW / ( 9.81 m/s²) × 206 m

m = ( 100 MW × 10⁶W/MW) / ( 9.81 m/s²) × 206 m

m = 10 × 10⁷ / 2020.86

m = 49484.37 kg/s

Therefore, the required mass flow rate is 49484.37 kg/s

5 0

2 years ago